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Bài 1:
A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + (\(\frac45\) - \(\frac{3}{17}\) + \(\frac13\)) - \(\frac17\) + (- \(\frac{14}{30}\))
A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + \(\frac45\) - \(\frac{3}{17}\) + \(\frac13\) - \(\frac17\) - \(\frac{14}{30}\)
A = (\(\frac15\) + \(\frac45\)) + (\(\frac{3}{17}\) - \(\frac{3}{17}\)) - (\(\frac43-\frac13\)) - \(\frac{30}{210}\) - \(\frac{98}{210}\)
A = 1 + 0 - 1 - (\(\frac{30}{210}+\frac{98}{210}\))
A = 1 - 1 - \(\frac{228}{210}\)
A = 0 - \(\frac{128}{210}\)
A = - \(\frac{64}{105}\)
Bài 2:
B= (\(\frac58\) - \(\frac{4}{12}\) + \(\frac32\)) - (\(\frac58\) + \(\frac{9}{13}\)) - (\(\frac{-3}{2}\)) + \(\frac{7}{-15}\)
B = \(\frac58\) - \(\frac{4}{12}\) + \(\frac32\) - \(\frac58\) - \(\frac{9}{13}\) + \(\frac32\) - \(\frac{7}{15}\)
B = (\(\frac58\) - \(\frac58\)) + (\(\frac32\) + \(\frac32\)) - (\(\frac13\) + \(\frac{9}{13}\) + \(\frac{7}{15}\))
B = 0 + 3 - (\(\frac{65}{195}\) + \(\frac{135}{195}\) + \(\frac{91}{195}\))
B = 3 - (\(\frac{200}{195}\) + \(\frac{91}{195}\))
B = 3 - \(\frac{97}{65}\)
B = \(\frac{195}{65}\) - \(\frac{97}{65}\)
B = \(\frac{98}{65}\)
\(\frac{3^6.45^4-15^{13}.5^{-9}}{27^4.25^3+45^6}\)=\(\frac{3^6.3^8.5^4-5^{13}.3^{13}.5^{-9}}{3^{12}.5^6+3^{12}.5^6}\)=\(\frac{3^{14}.5^4-5^4.3^{13}}{3^{12}.5^6+3^{12}.5^6}\)=\(\frac{3.1.}{1.5^2.}\)=\(\frac{3}{25}\)
Học tốt
a) 78 + 77 - 76 = 76(72 + 7 - 1)= 76.55 chia hết cho 55 (ghi kí hiệu, tại mk chưa pit đánh nó ở đâu xin lỗi)
b) 165 + 215 = 410 + 215 = (22)10 + 215
= 220 + 215
= 215(25 + 1) = 215.33 chia hết cho 33 (ghi kí hiệu nhé)
c) 817 - 279 - 913 = 328 - 327 - 326
= 326(32 - 3 - 1)
= 326.5= 322.34.5
= 322.405 chia hết 405
\(7^6+7^5-7^4\)
\(=7^4\cdot7^2+7^5\cdot7-7^4\)
\(=7^4\cdot\left(7^2+7-1\right)\)
\(=7^4\cdot55\)
\(=7^4\cdot5\cdot11⋮11\left(đpcm\right)\)
\(7^6+7^5-7^4=7^4.\left(7^2+7-1\right)\)
\(=7^4.55⋮11\)
\(=>7^6+7^5-7^4⋮11\)
\(a,7^6+7^5-7^4⋮55\)
\(7^4\left(7^2+7-1\right)⋮55\)
\(7^4\times55⋮55\left(dpcm\right)\)
\(8^{12}-2^{33}-2^{30}\)
\(=8^{12}-\left(2^3\right)^{11}-\left(2^3\right)^{10}\)
\(=8^{12}-8^{11}-8^{10}\)
\(=8^{10}\left(8^2-8-1\right)\)
\(=8^{10}\times55⋮55\left(dpcm\right)\)
a) ta có : \(7^6+7^5-7^4=7^4\left(7^2+7-1\right)=7^4.\left(49+7-1\right)=7^4.55⋮55\)
\(\Rightarrow7^4.55\) chia hết cho \(55\) \(\Leftrightarrow7^6+7^5-7^4\) chia hết cho \(55\)
vậy \(7^6+7^5-7^4\) chia hết cho \(55\) (đpcm)
b) ta có \(16^5+2^{15}=\left(2^4\right)^5+2^{15}=2^{20}+2^{15}=2^{15}\left(2^5+1\right)=2^{15}.\left(32+1\right)=2^{15}.33⋮33\)
\(\Rightarrow2^{15}.33\) chia hết cho \(33\) \(\Leftrightarrow16^5+2^{15}\) chia hết cho \(33\)
vậy \(16^5+2^{15}\) chia hết cho \(33\) (đpcm)
c) ta có \(81^7-27^9-9^{13}=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}=3^{28}-3^{27}-3^{26}\)
\(=3^{22}\left(3^6-3^5-3^4\right)=3^{22}\left(729-243-81\right)=3^{22}.405⋮405\)
\(\Rightarrow3^{22}.405\) chia hết cho \(405\) \(\Leftrightarrow81^7-27^9-9^{13}\) chia hết cho \(405\)
vậy \(81^7-27^9-9^{13}\) chia hết cho \(405\) (đpcm)
\(a.\)
\(7^6+7^5-7^4=7^4\left(7^2+7-1\right)=7^4.55⋮55\)
\(b.\)
\(16^5+2^{15}=2^{20}+2^{15}=2^{15}\left(2^5+1\right)=2^{15}.33⋮33\)
\(c.\)
Ta có : \(405=3^4.5\)
\(\Rightarrow81^7-27^9-9^{13}=3^{28}-3^{27}-3^{26}=3^{26}\left(3^2-3-1\right)=3^{26}.5⋮405\)
d) \(81^7-27^9-9^{13}\)
\(=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}\)
\(=3^{28}-3^{27}-3^{26}\)
\(=3^{26}\left(3^2-3-1\right)=3^{226}.5=3^{222}.405⋮405\)
b) 817 - 279 -913 chia hết cho 405
Ta có: 817 - 279 -913 = 328- 327-326
= 326(32-3-1)
= 326. 5 = 322. 405 chia hết cho 405 (đpcm)
a)
\(\frac{2}{5} - \left(\right. \frac{6}{7} + \frac{2}{5} \left.\right) = \frac{2}{5} - \frac{2}{5} - \frac{6}{7} = - \frac{6}{7}\)
b)
\(- \frac{5}{7} \cdot \left(\right. - \frac{4}{13} \left.\right) + \left(\right. - \frac{5}{7} \left.\right) \cdot \left(\right. - \frac{9}{13} \left.\right) = \frac{20}{91} + \frac{45}{91} = \frac{65}{91}\)
c)
\(\frac{27}{16} - \frac{14}{27} - \frac{22}{32} - \frac{13}{27} = \frac{27}{16} - \frac{14 + 13}{27} - \frac{11}{16} = \frac{27 - 11}{16} - \frac{27}{27} = \frac{16}{16} - 1 = 0\)
d)
\(\frac{3^{5} \cdot 8^{3}}{9^{2} \cdot 2^{8}} = \frac{3^{5} \cdot \left(\right. 2^{3} \left.\right)^{3}}{\left(\right. 3^{2} \left.\right)^{2} \cdot 2^{8}} = \frac{3^{5} \cdot 2^{9}}{3^{4} \cdot 2^{8}} = 3^{5 - 4} \cdot 2^{9 - 8} = 3 \cdot 2 = 6\)
a)\(\frac25-\left(\frac67+\frac25\right)\)
=\(\frac25_{}-\frac67-\frac25\)
\(=\left(\frac25-\frac25\right)-\frac67\)
\(=0-\frac67=-\frac67\)
b)\(\frac{-5}{7}\times\frac{-4}{13}+\frac{-5}{7}\times\frac{-9}{13}\)
\(=\frac{-5}{7}\times\left(\frac{-4}{13}+\frac{-9}{13}\right)\)
\(=\frac{-5}{7}\times\left(-1\right)=\frac57\)
C)\(\frac{27}{16}-\frac{4}{27}-\frac{22}{32}-\frac{13}{27}\)
\(=\frac{27}{16}-\frac{22}{32}-\frac{13}{27}-\frac{4}{27}\)
=\(\left(\frac{27}{16}-\frac{22}{32}\right)-\left(\frac{13}{27}+\frac{4}{27}\right)\)
\(=\left(\frac{54}{32}-\frac{22}{32}\right)-\frac{17}{27}\) \(=1-\frac{17}{27}=\frac{10}{27}\)
D)3^5.8^3/9^2.2^8
=3^5.(2^3)^3/(3^2)^2.2^8
=3^5.2^9/3^4.2^8
=3^4.3.2^8.2/3^4.2^8
=3.2=6